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xeze [42]
2 years ago
10

On her trips from home to school, karla drives along the streets after exiting the driveway. She drives 1.85 miles south, 2.43 m

iles east and 0.35 miles north. determine the magnitude of karla's resultant displacment. ​
Physics
1 answer:
12345 [234]2 years ago
6 0
0019235829mamNabshwwhbw
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ANSWER ASAP PLEASE 15 POINTS
Sholpan [36]

Answer:

SO4

Explanation:

I'm not sure if you're asking for more than one answer but S04 is one of them, you can search it up if you don't believe me

ion with the subscript of the atoms in a polyatomic ion. There is only ONE Cu and ONE S04, so get the charge for the Cu based on the S04. The formula is S04 , and there is only ONE S04 2, so Cu's charge here must be +2 for the compound to have an overall charge of zero. = copper (Il) ion S04 = sulfate ion then CuS04 = copper (Il) sulfate

3 0
3 years ago
Un pintor de 75.0 kg sube por una escalera de 2.75 m que está inclinada contra una pared vertical. La escalera forma un ángulo d
dezoksy [38]

Answer:

Work done, W = 1786.17J

Explanation:

The question says "A 75.0-kg painter climbs a 2.75-m ladder that is leaning against a vertical wall. The ladder makes an angle of 30.0 ° with the wall. How much work (in Joules) does gravity do on the painter? "

Mass of a painter, m = 75 kg

He climbs 2.75-m ladder that is leaning against a vertical wall.

The ladder makes an angle of 30 degrees with the wall.

We need to find the work done by the gravity on the painter.

The angle between the weight of the painter and the displacement is :

θ = 180 - 30

= 150°

The work done by the gravity is given by :

W=Fd\cos\theta\\\\=75\times 10\times 2.75\times \cos30\\\\=1786.17\ J

Hence, the required work done is 1786.17 J.

6 0
2 years ago
True or False:
bulgar [2K]

Answer:

False

True

True

True

True

False

8 0
3 years ago
How can you show that pressure increases with the increase of force?​
Julli [10]

Answer:

...........................

7 0
3 years ago
Read 2 more answers
A 3-kg wheel with a radius of 35 cm is spinning in the horizontal plane about a vertical axis through its center at 800 rev/s. A
Kruka [31]

Answer:

\omega_f = 585.37 \ rev/s

Explanation:

given,

mass of wheel(M) = 3 Kg

radius(r) = 35 cm

revolution (ω_i)=  800 rev/s

mass (m)= 1.1 Kg

I_{wheel} = Mr²

when mass attached at the edge

I' = Mr² + mr²

using conservation of angular momentum

I \omega_i = I' \omega_f

(Mr^2) \times 800 = ( M r^2 + m r^2) \omega_f

M\times 800 = ( M + m )\omega_f

3\times 800 = (3+1.1)\times \omega_f

2400 = (4.1)\times \omega_f

\omega_f = 585.37 \ rev/s

3 0
3 years ago
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